Question
Ā `sqrt(sqrt(20+sqrt(20+sqrt(20)) ... prop)` Ā =
?Solution
let x = `sqrt(sqrt(20+sqrt(20+sqrt(20)) ... prop)` So x = ā20+x So x2 = 20+x or x2-x-20 =0 or x = -4,5. x cannot be negative so final answer is 5. Alternate Method : For these type of questions of positive infinite, just brake the number in the product of two consecutive number and answer will be bigger one. 20 = 4 `xx` 5 so answer is 5. If the given number cannot be break in two consecutiveĀ numbers then for x= Ā `sqrt(sqrt(N+sqrt(N+sqrt(N)) ... prop)` , use this shortcut ,Ā x =Ā `{1 + sqrt(4N+1)}/2`
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